<p>The plane \(x + 2y - z = 4\) cuts the sphere \(x^2 + y^2 + z^2 - x + z - 2 = 0\) in a circle of radius</p>
Step-by-Step Solution
Key Concept: Find the sphere's center and radius using completing the square, then use the perpendicular distance formula from center to plane. The radius of the circle of intersection is √(R² - d²), where R is sphere's radius and d is the perpendicular distance from center to plane.
Step 1: Rewrite sphere in standard form by completing the square x^2 + y^2 + z^2 - x + z - 2 = 0 (x^2 - x) + y^2 + (z^2 + z) = 2 (x - 1/2)^2 - 1/4 + y^2 + (z + 1/2)^2 - 1/4 = 2 (x - 1/2)^2 + y^2 + (z + 1/2)^2 = 2 + 1/2 = 5/2 Center: C = (1/2, 0, -1/2), Radius: R = √(5/2) Step 2: Find perpendicular distance from C to plane x + 2y - z = 4 d = |1(1/2) + 2(0) - 1(-1/2) - 4|/√(1^2 + 2^2 + (-1)^2) d = |1/2 + 1/2 - 4|/√6 = |-3|/√6 = 3/√6 = √(3/2) Step 3: Apply circle radius formula r^2 = R^2 - d^2 = 5/2 - 3/2 = 1 r = 1 ∴ Answer: B
Correct Answer: B